MDoubt No 000007 - GATE

GATE [EE, 2014, 2 marks]
Question : To evaluate the double integral $\displaystyle\int\limits_0^8\left(\int\limits_{y/2}^{y/2+1}\left(\frac{2x-y}{2}\right)dx\right)dy$, we make the subsitution $\displaystyle u=\left(\frac{2x-y}{2}\right)$ and $\displaystyle v=\frac{y}{2}$. The integral will reduce to

(a) $\displaystyle\int\limits_0^4\left(\int\limits_0^2 2u du\right)dv$
(b) $\displaystyle\int\limits_0^4\left(\int\limits_0^1 2u du\right)dv$
(c) $\displaystyle\int\limits_0^4\left(\int\limits_0^1 u du\right)dv$
(d) $\displaystyle\int\limits_0^4\left(\int\limits_0^2 u du\right)dv$

Solution : Given $\displaystyle\int\limits_0^8\left(\int\limits_{y/2}^{y/2+1}\left(\frac{2x-y}{2}\right)dx\right)dy$
$u=\left(\dfrac{2x-y}{2}\right)$
$\implies du=dx$
$x=y/2 \implies u=0$
$x=y/2+1 \implies u=1$
$v=\dfrac{y}{2}$
$\implies dv=\dfrac{dy}{2}$
$x=0\implies v=0$
$x=0\implies v=4$
$\therefore$ the integral reduces to
$\displaystyle\int\limits_0^4\left(\int\limits_0^1 2u du\right)dv$
CORRECT ANSWER : B